of 1 71
A Genesis Project Based On A Six-fold Theory of Reality
By
Ian Beardsley
Copyright © 2023 by Ian Beardsley!
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Table of Contents
Abstract……………………………………………………………….3
Forward……………………………………………………………….4
1.0 Bridging the Microcosmos to the Macrocosmos…………5
2.0 Inertia As Proton-Seconds…………………………………….8
3.0 The Proton Radius………………………………………………13
4.0 Predicting The Life Elements………………………………….16
5.0 The Age of the Universe………………………………………..20
6.0 Summary…………………………………………………………..23
7.0 Overview…………………………………………………………..24
8.0 The Second Is The Metric………………………………………30
9.0 Kilogram-Seconds……………………………………………….35
10.0 Apophis……………………………………………………………39
11.0 Solving The System……………………………………………..42
12.0 The Proton Charge………………………………………………46
13.0 The Solar Magnetic Field……………………………………….48
14.0 Eightfold Symmetry……………………………………………..60
Appendix 1 Kinetic Energies Moon and Earth…………………….64
Appendix 2 Van Der Waals Radius…………………………………..65
Appendix 3 Chandrasekhar Limit……………………………………68!
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Abstract
We suggest the idea of a Genesis Project based on a six-fold theory of reality. Such a project
would be to solve the mystery of life-bearing worlds and systems so as to maintain them or,
even to transform dead worlds into living systems."
We develop the concept of intermediary mass somewhere between that of a proton and the
lower limit for a white dwarf star. From that we develop a constant k that bridges the
microcosmos with the macrocosmos. With this constant we predict the radius of a proton then
look at how it explains elements as based around 6-fold symmetry in the core life elements
carbon and hydrogen as a consequence of a theory for space, matter and time."
We find that though the Ancients developed the second by dividing the the Earth rotation and
orbit, and the Moon’s orbit into a convenient calendar, they unintentionally created the duration
of second we have today in a way that conveniently describe the atom and the Earth-Moon-
Sun System. It conveniently takes the unit of one in terms them making computing easier in
our model. As such we suggest the second is a Natural Metric."
Our six-fold theory of reality for a Genesis Project predicts the life span of the Universe, the
length of the Earth day, the length of the Earth year, and explains the ratio of the Moon’s kinetic
energy to that of the Earth in terms of the ratio of the radius of a proton to its mass. At the end
we suggest the magnetic field of the Sun may predict carbon, the core element of life
suggesting, it might play a role in a Genesis Project."
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Foreward
One of things I wanted to go into in this paper but didn’t is that you can speak of the structure of
the solar system even though it changes with time. This is important to understand when I refer
to sizes the Moon and the planets, and their orbital distances.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets. When I speak of the state of the solar system I speak of this point toward which the
solar system has formed and not the small changes that happen over millions of years due to
mutual interference between the bodies. In fact, mutual interference has torn apart possible
forming planets resulting in the current distribution we have today, because the current
distribution is more or less stable. The asteroid belt is a good example of this — it is a location
where a planet cannot form due to harmonic (repetitive) action on the orbital period at its
distance by orbital periods of planets beyond it. In short we take the state of the solar system as
an inflection point between what it became, and what it might minutely be going away from in
billions of years after it dies and can no longer support life.
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1.0 Bridging the Microcosmos to the Macrocosmos
To bridge the microcosmos to the macrocosmos we introduce a constant k."
Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational
Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where
’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
This follows from what Warren Giordano noticed that
Equation 1.0
We can eliminate and introduce the 6 of six-fold symmetry by introducing Avogadro’s
Number
We make an equation of state for the periodic table:
We can say for any element
Where is the number of protons in the element, so for carbon
Thus by equation 1.0 we have
Equation 1.1.
h
1 + α
α
10
23
10
23
N
A
= 6E 23
= 1
gra m
atom
N
A
𝔼 = 6E 23
𝔼
N
A
=
Z 6E 23pr oton s
Z gra m s
𝔼 =
Z gra m s
Z pr oton s
Z
=
6gra m s
6proton s
N
A
=
6(6E 23proton s)
6gra m s
h
(1 + α)
G
N
A
𝔼 = 6.0003kg
2
s
m
of 6 71
In the next step to bridge the microcosmos to the macrocosmos we create an intermediary
mass midway between the proton mass and the lower limit of mass for a white dwarf
star to form without collapsing into a blackhole star. It balances with its gravity by radiation
pressure alone, the so-called Chandrasekhar limit (See Appendix 3 for how it is derived):"
Equation 1.2
We make the approximation and define with the geometric mean:
Equation 1.3.
Giving
Equation 1.4.
Thus by equation 1.1 and equation 1.4 and using , we have
Equation 1.5.
We had
We can hone this by reintroducing 0.77 for 3/4
Thus precisely (Values of constants on page 8):
We have
We have honing our
m
i
m
p
M 0.77
c
3
3
G
3
N
m
4
p
= 1.41
0.77 3/4
m
i
m
i
= Mm
p
m
i
=
3
2
c
3
3
G
3
m
2
p
1/2
= h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
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Thus since we said with our estimate
We have a honed value of
This is our constant for bridging the microcosmos with the macrocosmos. We define the Earth
to be the ground state for the solar system yielding our six-fold basis for Nature
Equation 1.6.
Where is the average orbital velocity of the Earth.
We note the ground-state for an electron in a hydrogen atom gives
Where is the potential energy of the electron and is its rest energy.
m
i
= (67.9943)(1.01328) = 68.897kg 69kg 70kg
k =
1
m
2
i
h
(1 + α)
G
N
A
𝔼 =
1
773.5
s
m
k =
1
(68.897kg)
2
(6.62607E 34)
(1.007299)
6.67408E 11
6.02E 23 = 0.001268291s /m
1
k
= 788.4626
m
s
k v
e
= 6
k v
e
=
29790m /s
788.46m /s
= 6.145748
v
e
α
2
=
U
e
m
e
c
2
U
e
m
e
c
2
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2.0 Inertia As Proton-Seconds
We describe reality as being based in the six-fold. I really think Nature employs 6-fold
symmetry because I think it is the most dynamic being the product of the two smallest prime
numbers 2 and 3. Those are the smallest factors down to which anything can be reduced, so that
is why I think 6-fold is the basis of Nature, that the basis has to be reduced to the smallest
factors. As such we have 2x3=6, 3x3=9, 2x9=18 and 3x6=18. The periodic table of elements is
periodic over 18 groups, which means when you count to 18 the elements start over making
groups where they have similar properties falling into the same groups. These groups are
determined by electron configurations, for instance carbon is in group 14 which means since it
wants to have noble gas electron configuration, it gains 4 electrons to be the same as the noble
gas in group 18 of the same period. 18-14=4 meaning it ionizes as C4- so it has 4 electrons to
combine with 4 hydrogen atoms that are each H+ or so we can have two H atoms that can
combine with one oxygen atom which is O2- because oxygen is in group 16 meaning 18-16=2,
and so forth making life possible. We now need to show that the fundamental particles that
build reality are based on sixfold symmetry for their mass, size, and charge and we want to do it
in terms of gravity on the macroscale, thus it has to use G the universal constant of gravitation
and Planck’s constant h, that quantizes energy on the microscale. I find I can do this as such:
The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 2.1
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 2.2
Which describes mass per meter over time, which is:
Equation 2.3
m
P
: 1.67262 × 10
27
kg
h : 6.62607 × 10
34
J s
r
p
: 0.833 × 10
15
m
G: 6.67408 × 10
11
N
m
2
kg
2
c : 299,792, 459m /s
α :1/137
q
p
= q
e
= 1.6022E 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
G = 6.674E 11N
m
2
s
2
= 6.674E 11
m
3
s
2
kg
h
Gc
=
kg s
m
1.82E 16
kg s
m
of 9 71
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
Equation 2.4
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 2.5
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 2.6
We take the square root to get meters:
Equation 2.7
We multiply that with the value we have in equation 2.4:
Equation 2.8
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 2.9
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 2.10
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 2.11
Equation 2.12
α
1
α
2
(1.82E 16kg s /m)(18,769) = 3.416E 12kg s /m
α
2
=
U
e
m
e
c
2
S
p
S
p
= 4π r
2
p
= 8.72E 30m
2
S
p
= 2.953E 15m
(2.953E 15m)(3.416E 12
kg s
m
) = 1.009E 26kg s
m
p
m
h
=
1E 26
6secon d s
= 1.67E 27kg
m
c
=
1E 26kg s
1secon d
= 1E 26kg = 6proton s = 6m
p
m
p
=
1E 26kg s
6secon d s
m
c
=
1E 26kg s
1secon d
of 10 71
It seems the duration of a second is natural. If it is, since it was formed by a calendar based on
reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon
orbital mechanics. I find it is, that (See Appendix 1):
Equation 2.13
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by
very small amounts over millions of years. The solar system has evolved towards this since the
explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive
the second in terms of something else. For now we have the mass of a proton as:
Equation 2.14
This way of looking at things is to say matter is that which has inertia. This means it resists
change in position with a force applied to it. The more of it, the more it resists a force. We
understand this from experience, but what is matter that it has inertia? In this analogy we are
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. We are defining inertia as proton-
seconds, the action of subatomic particles over time. Our reasoning above in one equation is:
K E
moon
K E
earth
(Ear th Da y) 1secon d
m
p
=
3r
p
18α
2
4πh
Gc
of 11 71
Equation 2.15
Saying 1 second gives 6 protons is carbon (C)
Equation 2.16
And six seconds gives 1 proton is hydrogen:
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting cancel with the bodies of these equations on the left, but
rather divide into them, which are in units of mass, giving us a number of protons. I say this is
the biological because as we shall see our equations are based on one second is 6 protons is
carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are
the skeletons of biological life. We see this is a mystery of six-fold symmetry based around
biological life in the following computer program I wrote and its output:"
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number.
We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
1
α
2
m
p
h 4πr
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h 4πr
2
p
Gc
= 1proton 6secon d s = h ydrogen(H )
m
p
1
α
2
m
p
h4πr
2
p
Gc
of 12 71
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We nd exactly our equation predicts the second as"
Equation 2.16. "
We said"
"
But that was using the average orbital velocities. We nd using aphelions and perihelions (See
Appendix 1)"
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352secon ds
KE
moon
KE
Earth
(Ear th Day) = 1.2second s
K E
moon
K E
earth
(Ear th Da y) = 1.08secon ds
of 13 71
3.0 The Proton Radius
Thus we have the radius of a proton is given by carbon by evaluating at one second:
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
Substitute for to get
Where is the Van Der Waals radius for a hydrogen atom (See Appendix 2). We have
now introduced the radius of a hydrogen atom . Our formulation of inertia as
proton seconds is a form of impulse. To change that to momentum we have to divide by a
second. This radius of the hydrogen atom is the Van Der Waals radius, which is the closest
distance between two hydrogen atoms noncovalently bound. It is 120 pm. Divide that by
where 1/k is our constant
And we find
We have our equation for the radius of a proton
We only need to multiply it by to have the right units, and we get!
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
t
6
=
1
α
2
m
p
h 4πr
2
p
Gc
t
6
=
r
p
α
2
m
p
h 4π
Gc
R
H
/2
r
p
t =
R
H
2α
2
m
p
h 4π
Gc
R
H
R
H
= 1.2E 10m
ck
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
1 + α
1
N
A
𝔼
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
N
A
𝔼
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12secon d s
3 2
16
h
Gπ
1
α
2
m
2
p
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
t /ck = 1secon d
of 14 71
Equation 3.1.
Then suggest we picked up 9/8 in approximations which is close to one anyway so we write
Equation 3.2.
We form constants:
And we have the Equation:
Equation 3.3
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
Equation 3.4.
If our equation is right and we put it into natural units then the product should be close to
one:
Let us start with the units with which we are working:
r
p
=
9
8
2
1
m
p
hc
4π
3
G
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
= 8.26935E 16m 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E 9kg
R
H
N
A
𝔼
= 1.99E 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
m
p
G =
m
3
kg s
2
h = kg
m
2
s
c = m /s
of 15 71
And convert these to proton-masses and proton-radii:
Now we find k in these units:
Thus we have:
Equation 3.5.
"
We will say our theory is holistic and that we are describing the proton radius in terms of the
whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van Der
Waals radius, which is determined by hydrogen gas, or (See Appendix 3 for the theory of the
Van Der Waals radius)
G = 6.67408E 11
m
3
kg s
2
1.67262E 27kg(0.833E 15m)
3
s = 193,131, 756
h = 6.62607E 34kg
m
2
s
s
(0.833E 15)
2
(1.67262E 27kg)
= 5.71E 23
c =
(299,792, 459m /s)(1sec)
(0.833E 15m)
= 3.6E 23
R
H
=
1.2E 10m
0.833E 15m
= 144,058
k =
hc
2π
3
G
= 6.93E 9kg
k =
(5.71E 23)(3.6E 23)
2π
3
(193131756)
= 4E18proton m a sses
r
p
m
p
= k
R
H
N
A
𝔼
r
p
m
p
=
(4E18)(144058)
(6E 23)
=
5.76E 23
(6E 23)
= 0.96 1
H
2
of 16 71
4.0 Predicting The Life Elements
We have that"
Equation 4.1 "
Equation 4.2. "
Which give the core life elements carbon and hydrogen, which are also the hydrocarbons that
make-up the skeletons of biological life chemistry:"
Equation 4.3. "
Saying one second gives 6 protons is carbon (C) and, six seconds gives 1 proton is hydrogen
(H):"
Equation 4.4. "
We have the six-fold symmetry in the constant k and the orbital velocity of Earth:"
Equation 4.5. "
We substitute Equation 4.2 and 4.5 into equation 4.3 giving:"
Equation 4.6"
"
Or we can let cancel with the numerator and not write in protons:"
Equation 4.7"
"
We see the radius of a proton to the mass of a proton is proportional to the kinetic energy of
the moon to the kinetic energy of the Earth, times the Earth rotation period (EarthDay) times the
square root of the Earth’s orbital velocity and equal by the constant square root k.!
1
6α
2
r
p
m
p
h4π
Gc
= 1secon d
KE
Earth
KE
moon
(Ear th Day) = (1secon d )
1
α
2
m
p
h4πr
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h4πr
2
p
Gc
= 1proton 6secon ds = hydrogen(H )
k v
e
= 6
1
α
2
r
p
m
p
h4π
Gc
KE
earth
KE
moon
1
Ear th Day
= k v
e
protons
m
p
1
α
2
r
p
m
p
h4π
Gc
= k
1/2
KE
moon
KE
earth
(Ear th Day) v
e
of 17 71
But the orbital velocity of the Earth is given by the Sun’s mass and Earth’s distance from the
Sun:"
Equation 4.8. "
But the Moon perfectly eclipses the Sun which means as seen from the earth the Moon
appears to be the same size as the Sun. This is because"
Equation 4.9. "
Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius.. Thus in terms of the moon and the Sun"
Equation 4.10. "
So we have from 4.8"
Equation 4.11. "
This means from 4.7"
Equation 4.12"
"
Interesting I found that the radius of the Sun to the Moon’s orbital radius in the ratio of a mass
of a gold atom to the mass of a silver atom which is about 1.8 or 9/5 by comparing the molar
mass of gold to that of silver. Interesting because gold and silver have been used since ancient
times as metals for making ceremonial jewelry and the Sun is gold in color and the Moon is
silver in color. We can write this"
Equation 4.13. "
Au (gold) and Ag (silver) are also some of the finest metals for electronic circuitry as they are
such fine conductors. The above equation 4.12 has the radius of a proton and its mass
described by the kinetic energy of the Moon to that of the Earth perhaps because the Moon
makes life possible holding the Earth at its inclination to its orbit around the Sun allowing for
the seasons, and this ratio might be optimal for that. Naturally to optimize the success of life on
Earth this might have to be a factor of the Earth’s rotation which is given above by EarthDay.
v
e
v
e
=
GM
r
e
r
e
r
m
R
R
m
r
e
r
m
R
R
m
r
e
=
R
R
m
r
m
v
e
= G
M
R
R
m
r
m
1
α
2
r
p
m
p
h4π
Gc
= k
1/2
KE
moon
KE
earth
(Ear th Day) G
M
R
R
m
r
m
1/2
R
r
m
=
Au
Ag
= 1.8
of 18 71
The orbital velocity of the Earth in the equation is determined by the Earth’s distance from the
Sun and the radius of a proton to the mass of a proton seems to determine that because
perhaps the velocity of the Earth determines its distance from the Sun and that in turn happens
to be the habitable zone, because it is the distance from the Sun where water can exist on
Earth in its liquid phase. Essentially equation 4.12 says how the kinetic energy of the Moon is in
the ratio of the radius of a proton to its mass, and it is that kinetic energy of the Moon that
makes like possible on Earth. Further the moon is composed of those same protons, and the
neutrons which have a similar ratio in their radii and mass. I would think this equation would be
central to any kind of “Genesis” project where we try to understand the secrets of making life-
bearing worlds so we can maintain or convert dead worlds like Mars to living worlds so we can
colonize them."
The length of the Earth day is related to the annual average temperature of the Earth, and as it
would turn out Mars, the one planet in our solar system we can colonize most eectively, has a
day of the same length as that of the Earth, 24 hours."
Interestingly we can speak of the day of Mars and the day of the Earth interchangeably
because they are about the same, about 24 hours long for their rotations. We have to keep in
mind that Mars is smaller than Earth regardless that it completes a rotation in the same time.
The radius of Mars is 3,389,500 meters and that of the Earth is 6,378,100 meters. That is at its
surface Mars rotates through 246.5 m/s and at the surface of the Earth 463.8 m/s."
But interestingly, this ratio in size of Earth to the size of Mars is"
Equation 4.14 "
And the ratio of the Mars year to the Earth year is the same"
Equation 4.15 "
Interesting because the size of a planet has nothing to do with its orbital period. Orbital period
is a function of the distance of the planet from the star, not of its size. You could have an
asteroid at Mars orbit much smaller than Mars. The mass of Mars, of course is much lower too.
The mass of Mars is 6.39E23 kg and that of Earth is 5.97219E24 kg giving:"
Equation 4.16 "
The Earth is then about 9 times more massive than Mars. Equations 4.14 and 4.15 are to say"
Equation 4.17 "
Where is the Earth radius and is the Mars radius, and is the Earth year and is the
Mars year. This is interesting because it is similar to the parameters of the moon with respect
to the Earth and Sun that we suggest is a mystery that is a message from unknown forces. We
had"
6,378,100
3,389,500
= 1.88
687
365.25
= 1.88
5.97219E 24
6.39E 23kg
= 9.34615
R
earth
R
mars
=
T
mars
T
earth
R
e
R
m
T
m
T
m
of 19 71
Equation 4.18 "
Where is the earth orbital radius, is the Moon’s orbital radius, is the Sun’s
radius, and is the radius of the Moon. Indeed the mystery of the Moon could speak to us
of the mystery of our existence even to our ancient ancestors flaking spearpoints from stone
around the fire a million years ago because they could witness the Moon perfectly eclipsing the
Sun, as much as it could be a message to us today. But in 15.4 where the Earth size to the
Mars size equals the Mars year (journey across the sky) to the Earth Year (journey of the Sun
across the sky) would require us to know the size of the Earth and the size of Mars, so if it is a
message like the lunar message, it would only be readable later in the human story."
Given Kepler’s Law of Planetary motion"
"
Then we have"
Equation 4.19. "
Bear in mind, interestingly, the ratio in size of Earth to the size of Mars is"
"
And the ratio of the Mars year to the Earth year is the same"
"
But as well and we will call it equation 4.20"
Equation 4.20. "
Where Au and Ag are the masses of gold and silver atoms. The 1.8 is close to 1.88."
r
earth
r
moon
=
R
R
moon
r
earth
r
moon
R
R
moon
T
2
GM
4π
2
= r
3
(
T
mars
R
mars
R
earth
)
2
GM
4π
2
=
(
r
moon
R
moon
R
)
3
6,378,100
3,389,500
= 1.88
687
365.25
= 1.88
R
r
moon
=
Au
Ag
= 1.8
of 20 71
5.0 The Age of the Universe
We have the possible core equation for a so-called Project Genesis giving the ratio of the radius
of a proton to its mass the kinetic Energy of the moon to that of the Earth:
We now see the same equations predict the age duration of the Universe in the standard model.
That is we start with the following two equations:
Eq 5.1.
Eq. 5.2
We want to factor 5.1 into a ratio between kinetic energies times a time as we have in 2.2. We
write 2.1 as
And noticing
Which gives
Is
And we know
And have
1
α
2
r
p
m
p
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Da y) k v
e
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
K E
moon
K E
earth
(Ear th Da y) = 1secon d
6α
2
r
p
m
4
p
Gc
4πh
=
m
p
1secon d
m
3
p
Gc
h
= K E
36α
4
m
p
KE
4π
=
r
2
p
m
2
p
1secon d
K E =
4π
36α
4
r
2
p
m
p
t
2
1
t
1
=
1
6α
2
m
p
h 4πr
2
p
Gc
of 21 71
We also notice that obviously by dimensional analysis
We have the kinetic energy and we need another kinetic energy to form the ratio:
We have the intermediary mass and the constant k inverted. We can write
This is
=3.03E56 seconds
But we also formulated
4π
36α
2
(0.833E 15)
2
t
2
1
(1.67262E 27) = 1.413E 49J
m
3
p
Gc
h
=
4π
36α
4
r
2
p
m
p
t
2
1
h
G
c
3
m
p
= K E
h
G
c
3
m
p
(K E
moon
)
= 4.6654
K E = m
3
p
Gc
h
m
3
p
Gc
h
K E
(t im e) = aT im e
m
i
K E = m
i
(
1
k
)
2
K E = (68.897kg)(788.4626m /s)
2
= 42,831,358Joules
K E
1
K E
2
(t im e) = (aT im e)
m
i
(
1
k
)
2
m
3
p
Gc
h
(1secon d ) =
(42,831,358J )
(1.4130E 49J )
(1secon d )
of 22 71
So we can write
1 Earth year = (365.25)(24)(60)(60)=31557600 seconds
The universe is theorized by standard models to die in 100 trillion years, which is when the last
stars born will die out. This is exactly 1E14 years
We see the moon is connected to how old the universe is theorized to become.
6.62607E 34
6.67408E 11
299792459
3
1.67262E 27
= 1.599298E 29J
h
G
c
3
m
p
= 1.599298E 29J
h
G
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) =
1.599298E 29J
42831358J
= 3.734E 21secon d s
3.734E 21s
31557600s
= 1.1832332E14years 1E14years
of 23 71
6.0 Summary
We would like to present some things clearly:
Thus
If is 1.00 seconds even. In terms of our theory
Or as another estimate:
Because
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
1
6α
2
m
p
h 4πr
2
p
Gc
= 1.004996352secon d s
t =
R
H
2α
2
m
p
h 4π
Gc
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
1 + α
1
N
A
𝔼
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
N
A
𝔼
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12secon d s
3 2
16
h
Gπ
1
α
2
m
2
p
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
t
1
=
R
H
π16α
2
m
2
p
3 2
N
A
𝔼
h
G
= 1.12secon d s
t
1
=
R
H
πα
2
m
2
p
2
6
1
N
A
𝔼
h
G
= 1.067secon d s
3 2
16
= 0.265165
2
6
= 0.23570
0.23570
0.265165
= 0.888888
of 24 71
7.0 Overview
If the universe was born from a big explosion called “the big bang” then it is expanding and the
distances between galaxies is growing. This does not mean there is an edge to the universe —
space could continue on infinitely in all directions but there could be a horizon beyond which
there is no content.
There can be an observable edge to the universe, but it is more of a horizon, just as there is a
horizon to the ocean, we know beyond the horizon there is more water. Since the universe was
born 13.82 billion years ago in the Big Bang we can only see those galaxies whose light has taken
less than 13.82 billion years to reach us. These galaxies form a sphere around the Earth called
“the observable universe”. It is 92 billion light years in diameter because in its beginning the
universe inflated much faster than the speed of light.
The Schwartzchild radius:
Gives the event horizon of a black hole. I have heard it said if the mass, M, is the mass of the
universe, that radius is the edge of the observable universe that, the observable edge of the
universe is actually the event horizon of a black hole and that then the Universe is a giant black
hole and that we are inside of it. I would like to call this to question, but first let’s discuss the
types of possible universes.
The pivotal parameter describing the universe is 𝛀 (omega) which is the average matter density
of the universe divided by a critical value of that density. Whether omega is less than1, 1, or
greater than 1 determines whether the Universe is open, flat, or closed. If matter is mostly inert
as in the dust models, there is a particular fate for each omega. Since 1998 the observations in
supernovas indicate the universe is accelerating in its expansion. To explain this cosmologists
hypothesize the existence of dark energy which can be any field with negative entropy causing
negative pressure. If 𝛀<1 then the Universe is open and the fate of the universe is in its heat
death, the burning out of the last stars born. This is also true of a flat universe if it has the
hypothesized dark energy, because then it accelerates to escape collapse. The fate of the universe
depends on its density and today most evidence points to it will expand indefinitely. In this
scenario there is enough a supply of gas for stars to form for 10¹² to 10¹⁴ years. 1 to 100 trillion
years. That is the last stars born would die in 10¹⁴ years. There are models that suggest the
universe is eternal where as stars die more matter comes into existence from which more stars
are continuously forming.
Now let’s call into question that the edge of the observable universe is the Schwartzchild radius,
the event horizon of a blackhole.
I had seen a documentary where they said there was a cosmologist working on a theory that the
edge of the observable universe is the event horizon of a blackhole. I thought okay as weird as
that seems, and I say weird because that means either the Earth is at the center of this blackhole,
or if not, mysteriously wherever you are in the universe is the center. I accepted this as possible
because the more we learn the stranger the universe gets. Indeed a long time ago Niels Bohr said
that if you haven’t realized that quantum mechanics says everything that makes up the universe
can’t be considered real, you have missed its point. And indeed if you think about quantum
mechanics, this rings true, so yes I was able to accept this about the edge of the universe, as
R
s
=
2GM
c
2
of 25 71
possible. However, taking a second look we see what is going on here may be circular
reasoning…
Einstein’s equations give the relationship between spacetime and matter content of the
Universe. From this Friedmann gave us his equations that are at the heart of cosmology, which
are functions of time, the scale-factor R, the matter density of the universe , and the pressure p
due to radiation produced by the stars, or by the galaxies because they contain the stars:
Where
for a 3-sphere, for 3-space, and for a 3-pseudosphere. These Friedmann
equations give the critical density of the universe as
Where is the Hubble constant which is the expansion rate of the universe. Current estimates
are that it is . A Hubble sphere is given, which is given by
,
Since the density of the universe is very close to the critical density which is
, then since
So this is close to the observable edge of the Universe. But the radius of the Universe is much
larger than 13.8 billion light years, because it is expanding we say it is 46.508 billion light years.
However we should look at where we got this mass we used, it came from the Hubble constant
and we used . What we did is:
ρ
2
R
R
+
(
R
R
)
2
+
κ
R
2
+ 8π p = 0
(
R
R
)
2
+
κ
R
2
=
8πρ
3
R
dR
dt
κ = + 1
κ = 0
κ = 1
ρ
c
=
3H
2
0
8π G
H
0
71k m /s /Mpc
R
HS
=
c
H
0
=
299792549m /s
71,000m /s /Mpc
= 4222.429Mpc
pc = 3.26l y
Mpc = 3.26 E6ly
R
HS
= 13.8E 9ly
ρ
c
8.6E 30g /cm
3
= 8.6E 27kg/m
3
R
HS
= 1.38E10l y = 1.3E 26m
M = ρ
c
4
3
π R
3
= 8E52kg
R
s
=
2GM
c
2
= 2
(6.67E 11)(8E52kg)
299792459
2
= 1.18742E 26m = 12.55E 9ly
ρ
c
of 26 71
Which is
Which is the critical mass given by the Friedmann equations. So, we have gone in a circle, I
think. Let us use our value of 46.508E9 ly = 4.4E26m instead of 1.3E36m:
That is larger by a factor of 36.255. Not to say we aren’t in a blackhole, it just can’t be at the edge
of the observable universe, the radius of the Hubble sphere. Now with this brief sketch of
cosmology that we have done, let us talk again about some of the conclusion in this paper. We
have the second may be a natural duration. We suggest this because we found"
"
Or…"
"
R
s
=
2GM
c
2
= 2
G
c
2
ρ
4
3
π
(
c
H
0
)
3
=
8π G
3H
2
0
ρ
c
R
ρ
c
=
3H
2
0
8π G
M = ρ
c
4
3
π R
3
= 3.0686E54kg
R
s
=
2GM
c
2
= 2
(6.67E 11)(3.0686E54kg)
299792459
2
= 4.55E 27m = 4.81E11l y
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
K E
moon
K E
earth
(Ear th Da y) = 1secon d
h
G
c
3
m
p
m
i
(
1
k
)
2
K E
moon
K E
earth
(Ear th Da y) = Li feSpa nUniverse
h
G
c
3
m
p
m
i
(
1
k
)
2
1
6α
2
r
p
m
p
h 4π
Gc
= Li feSpa nUniverse
t
1
=
R
H
π16α
2
m
2
p
3 2
N
A
𝔼
h
G
= 1.12secon ds
t
1
=
R
H
πα
2
m
2
p
2
6
1
N
A
𝔼
h
G
= 1.067secon ds
of 27 71
But let us ask why did the unit of a second come out as natural in the structure of Nature and
the Universe. We got it from ancient times by dividing the Earth day (1 rotation of the Earth) into
24 hours and that into 60 minutes and that into 60 seconds which is to divide it into"
"
"
The smallest primes, smallest factors down to which we can reduce a fraction are, 2 and 3. We
have"
, "
And there are 18 groups in the periodic table of elements because the properties of the
element are periodic over cycles of 18. We have"
, "
, "
"
Is the golden ratio."
And degrees come from in circle giving"
"
Is unity."
"
"
"
"
(24hours)(60 minutes)(60second s) = 86400secon ds
1Rotation
86400s
=
1Day
86400s
2 3 = 6
3 6 = 18
86400
2
= 43200
86400
3
= 28800
86400
12
= 7200
86400
6
= 14400
7200
100
=
7200
10
2
= 72
2 cos(72
) = Φ =
5 + 1
2
360
86400
360
= 240
2 cos(240
) = 1
86400
6
= 14400
14400
8
= 1800
86400
8
= 10800
6
8
=
3
4
= 0.75
of 28 71
So we are dealing with the reconciliation of the six-fold with the eight-fold. This suggests to me
the periodic table of the elements, and I had found something in earlier work that I had done
this is group 14 in the periodic table of elements its first two elements carbon, the core element
of life and silicon (Si) the core element of Earth (or sand) which is silicon dioxide ( ). Let’s
look at this on the next page…."
SiO
2
of 29 71
We see that the atomic radius of silicon the core element of artificial intelligence (transistor
technology) fits together with the core element of biological life carbon if the silicon is taken as
inscribed in a regular dodecagon (12 sides) and the carbon is taken as inscribed in a regular
octagon (8 sides). We have:"
, , "
, , "
Apothem: "
For a regular dodecagon:"
"
The radius of a silicon atom is Si=0.118nm and that of carbon is
C=0.077nm:"
"
"
"
This has an accuracy %"
Twelve is an expression of six in that it allows you describe it as two three’s and three two’s.!
D = 1 + 2x
x
2
+ x
2
= 1
2
2x
2
= 1
2x
2
= 1
x = 2 /2
D = 1 + 2
a = (1 + 2)/2 = 1.2071
a =
s/2
tan(θ /2)
=
0.5
tan(15
)
= 1.866
Si
C
=
0.118
0.077
= 1.532
a
S
i
a
C
=
1.866
1.2071
= 1.54585
1.532
1.54585
100 = 99
of 30 71
8.0 The Second Is The Metric
We have "
8.1. "
8.2.
, and . In both 11.1 and 11.2 we have
kinetic energy over kinetic energy times one second equal to a pivotal time period in Nature, the
first the Life Span of the Universe, and the second the Earth Day, period of rotation of the Earth.
In equation 11.1 is given by the white dwarf star:
And
Is the Chandrasekhar limit for the mass of a star to not collapse into a blackhole so it can
become a white dwarf star. The Life Span of the Universe is perhaps related to the Schwartzchild
radius which gives the event horizon of a blackhole star. It is
8.3.
If is the mass of the Universe
8.4.
Where is the radius of the Universe and is the critical density of the Universe that
determines whether the universe is open, flat, or closed. It is
8.5.
Where
h
G
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) = Li feSpanUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
c
3
m
p
= Kin et icEn erg y
m
i
(
1
k
)
2
= Kin et icEn erg y
m
i
m
i
= Mm
p
M = 0.77
c
3
h
3
8π
3
G
3
m
4
p
R
s
=
2GM
c
2
M
M =
4
3
π R
3
ρ
c
R
ρ
c
ρ
c
=
3H
2
0
8π G
of 31 71
8.6.
Is the radius of the Hubble sphere is observable edge of the Universe. is the Hubble constant
is the expansion rate of the Universe which is thought to be approximately 71km/s/Mpc. And,
the critical density is approximately the average density of the Universe. We can write 11.1
8.7.
Using where .
We have
EarthDay=(24)(60)(60)=86400 seconds
LifeSpanUniverse=1E14 Years = 3.734E21 seconds
KE Moon=3.428E28J
KE Earth=2.7396E33J
Kinetic energies used from aphelions and perihelions calculated earlier in this paper. We have
=
Using our honed value for k.
=(5.354E-34)(1.5657E-10)=8.3827578E-44
Multiplying these last two:
(3.96631kg)(8.3827578E-44)=3.32486E30 kg
This mass should equal that of a white dwarf star (Its lower limit not to collapse into a
blackhole). There can be some play as we are suggesting the moon is an indicator somehow of all
that is going on in the Universe, and it only has to be at approximately the right orbit to do this.
We have
Which makes the equation pretty accurate in this form considering we are comparing something
like the Earth day to something as immense as the LifeSpanUniverse. We are going to want to
look at:
c
H
0
= R
HS
H
0
ρ
c
ρ
k
4
m
3
p
h
2
c
6
G
2
(
Ear th Da y
Li feSpa nUniverse
)
2
(
K E
moon
K E
earth
)
2
= M
WhiteDwar f
m
i
= Mm
p
M = M
WhiteDwar f
k
4
m
3
p
h
2
c
6
G
2
=
(
1
788.4626
)
4
1
(1.6726E 27)
3
(6.626E 34)
(6.674E 11)
2
(299792459)
6
(2.58747E 12)(2.1371E 80)(9.926581E 47)(7.25979E50) = 3.96631k ilogra m s
(
Ear th Da y
Li feSpa nUniverse
)
2
(
K E
moon
K E
earth
)
=
(
86400
3.734E 21
)
2
(
3.428E 28J
2.7396E 33J
)
2
2.86416E 30kg
3.32486E 30kg
= 86 %
of 32 71
Where is the Chandrasekhar limit. We now write
Where is the Schwartzchild radius is the is the edge of the observable universe. If
we have
Then
Yields
Which is the critical density of the Universe. We write 8.8
8.8.
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
8π G
3H
2
0
(
c
H
0
)
ρ
c
= R
s
R
s
c
H
0
= R
HS
R
HS
= R
S
8π G
3H
2
0
(
R
s
)
ρ
c
= R
s
8π G
3H
2
0
ρ
c
= 1
ρ
c
=
3H
2
0
8π G
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
H
0
= 71,000m /s /Mpc
(3.26E6ly /Mpc)(9.461E15m /ly) = 3.1E 22m /Mpc
H
0
= (71000m /s)(3.1E 22m) = 2.3E 18s
1
ρ
c
=
3(2.3E 18)
2
8(3.141)(6.6741E 11)
= 9.461E 27kg/m
3
of 33 71
Thus we have
Where is a white dwarf star. We have
We suggest for some mass , we have
It is given by
8.9
For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.
Conclusion next page…
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
M
h
G
c
3
m
p
M
(
1
k
)
2
(1secon d ) = 1Ear thYear
M =
h
G
c
3
m
p
(1Ear thYear)
(
1
k
)
2
(1secon d )
h
G
c
3
m
p
=
6.62607E 27
6.67408E 11
(299792459)
3
1.67262E 27
= 1.599E 36J 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
M =
1.599E 36
621673
1
31557600
= 8.15E 22kg
8.15E 22kg
7/34767E 22kg
= 1.10919516 1.12m oon s
of 34 71
Thus we have
Where is a white dwarf star and gives LifeSpanUniverse.
Where Earth-Moon-Sun orbital parameters give the Earth rotation period.
Where is the mass of the moon gives the Earth orbital period. We have already formulated
From the radius of a proton and the mass of proton , our theory panned out: The second is
a unit of time that expresses the cycles of the Universe, it is the metric (measures thing
conveniently). I think next we want to look at
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
M
m
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
r
p
m
p
8π G
3H
2
0
(
c
H
0
)
ρ
c
= R
s
c
H
0
= R
HS
ρ
c
=
3H
2
0
8π G
of 35 71
9.0 Kilogram-Seconds
One of the things we have done is said since the Chandrasekhar limit can be taken as 1.44
solar masses, and the mass of the Sun is 1.98847E30kg, and the mass of a proton is
1.67262E-27kg, that our intermediary mass is"
"
We also computed"
Where came from the approximation of (3/4) for 0.77 in
Which we ultimately took as our value for the honed :
We have
We have honing our
We developed a concept of proton-seconds. Now we develop a concept of kilogram-seconds. To
do this we look at our equations for the cycles of the Universe:
9.1.
9.2.
m
i
= (1.44)(1.98847E 30kg)(1.67262E 27) = 69.205kg
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3/2
m
i
= Mm
p
= 0.77
c
3
3
G
3
N
m
2
p
1/2
m
i
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg 69kg 70kg
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
of 36 71
9.3.
9.4.
We consider
9.5.
9.6.
And write
9.7.
Which is
9.8.
Where we remember
9.9.
Can be taken as
9.10.
If cancels with the numerator.
9.11.
9.12.
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
h
G
c
3
m
p
=
6.62607E 27
6.67408E 11
(299792459)
3
1.67262E 27
= 1.599E 36J 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
h
G
c
3
m
p
(
1
k
)
2
(1secon d ) = 2.5721E 30kilogr am secon ds
h
G
c
3
m
p
t
1
k
2
= 2.5721E 30kg s
1
6α
2
r
p
m
p
h 4π
Gc
= proton secon ds
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
m
p
h
G
c
3
m
p
1
6α
2
r
p
m
p
k
2
h 4π
Gc
= 2.5721E 30kilogr am secon ds
1
6α
2
r
p
m
p
h 4π
Gc
= proton secon ds
of 37 71
We can now take 9.8 and divide by the mass of the moon:
(365.25)(24)(60)(60)=3157600 seconds is:
Meaning we have the Earth year to 90% accuracy. We can also divide by the mass of the Earth:
=
And we note as we said in the beginning of this paper the moon orbits the earth inclined 5
degrees to the Earth orbital plane (That is, to the ecliptic). In days the Earth goes about 5
degrees around the Sun because the earth goes through about 1 degree a day because there are
360 degrees in a circle and 365.25 days in a year. We can also divide kilogram-seconds by the
mass of the Sun to get:
And find it is very close to one second. We in fact have that if we use instead of aphelions and
perihelions to calculate the kinetic energies of the moon and earth, we can use their average
orbital velocities we have (Appendix 1):
Equation 9.13.
As opposed to
Equation 9.14.
Which allows us to write very accurately
h
G
c
3
m
p
t
1
k
2
= 2.5721E 30kg s
2.5721E 30kg
M
moon
=
2.5721E 30kg s
7.34767E 22kg
= 35005654.85s
31557600s
35005654.85
= 90 %
2.5721E 30kg
M
earth
=
2.5721E 30kg s
5.972E 24kg
= 430693.2351s
430693.235s
31557600s
= 0.013647845years
73.27
c ycles
year
365.25d ays
73.27
= 4.98498d ays 5d ays
2.5721E 30kg
M
=
2.5721E 30kg s
1.989E 30kg
= 1.29316s
K E
moon
K E
earth
(Ear th Da y) = 1.2secon ds
K E
moon
K E
earth
(Ear th Da y) = 1.08secon ds
of 38 71
Equation 9.15.
As we can see while kilogram-seconds deals with Natural cycles and periods
Proton-seconds are structured around the core element of life, carbon, and the most abundant
elements in life carbon and hydrogen and in the most abundant element in the Universe,
hydrogen, as founded on six-fold symmetry characterized in the hydrocarbons, the skeletons of
biological life and organic chemistry:
"
h
G
c
3
m
p
t
1
M
k
2
=
K E
moon
K E
earth
(Ear th Da y)
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
1
α
2
m
p
h4πr
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h4πr
2
p
Gc
= 1proton 6secon ds = hydrogen(H )
of 39 71
10.0 Apophis
The reason that motivated me to write this paper was explained in the opening section, but not
in its more specific details. It was explained that the Moon seems to give us a hint that we are
here for a reason, like it perfectly eclipses the Sun, and that may have been a message to our
ancient ancestors, as well as to us today. Or in that the Moon holds the Earth’s inclination to
the ecliptic, that keeps the climate right to support life. These things were already known, but I
discovered something that made me pursue this as a paper, and that was in that both the
proton and the Moon define that duration of a second nearly perfectly, which I have found in
many ways make it a good unit for measuring Nature accurately. I call it a “Natural Metric”. In
light of how we got the second makes this an even deeper mystery; We got it from the way the
ancients divided up the rotation of the Earth and the Earth’s orbit to make a calendar that
works well. But I pursued this paper for an even deeper reason than mentioned: I was
wondering in light of negative forces in the world and positive forces in the world whether or
not we were slated for success as a species from the outset, and the Moon in light of what we
already knew, and what I found, gave me good reason to pursue the project. It panned out so
well that I began to feel there was something more than certainly going on here. Among other
things, like a theory for reality, I found what seems to be life cycles of the Earth and Universe. I
found"
"
h
G
c
3
m
p
t
1
M
k
2
=
K E
moon
K E
earth
(Ear th Da y)
h
G
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
1
α
2
m
p
h4πr
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h4πr
2
p
Gc
= 1proton 6secon ds = hydrogen(H )
of 40 71
I decided to apply this to Apophis, an asteroid we thought might hit Earth in 2029 that could
have either wreaked havoc, or even brought about our extinction. Hence its name from the
Greek meaning “mayhem” and “chaos”. It is going to be in 2023 the nearest miss in history,
coming between us and the Moon and even within the orbits of our geosynchronous satellites. It
has an orbit of about year and comes near to Earth every seven years. We want to see if the
Moon predicts such an asteroid (of 7 years) with equation 12.3. This gives
The mass of the Moon is
But is predicted by equation 12.3 to be 8.15E22kg. Thus we divide this by our answer above for
M:
This may be the 7 years, where the year is one orbit of the Earth and as well approximately that
of Apophis. Much more needs to be done, but we seem to be doing well with these equations of
Natural cycles so far. Of course with the second as a Natural Metric, the year is Natural as well
so the easiest way to do this is in equation 12.3 since one lunar mass gives 1 Earth year, so will
the 7 years of Apophis near miss give seven lunar masses. But considering the system of
equations given above, the question becomes why 7 years gives 7 lunar masses. We have to solve
this system of equation above, further. We only know Apophis passes between the Earth and the
Moon in 2023. In this paper I offer it up with more editing and correction of typos, and the
above suggestion of looking further into the system of equations for Apophis. This makes us
suggest mass is inversely proportional to time:
M =
h
G
c
3
m
p
(1Ear thYear)
(
1
k
)
2
(1secon d )
h
G
c
3
m
p
=
6.62607E 27
6.67408E 11
(299792459)
3
1.67262E 27
= 1.599E 36J 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
7Ear thYear = 220903200s
(1.599E 36J )
(220903200s)(621673.27)
= 1.16435E 22kg
M
m
= 7.34767E 22kg
Apophis =
8.15E 22
1.16435E 22kg
= 6.9996 7
T =
1
M
of 41 71
Where T is in Earth years and M is in Lunar masses. This is a good clue for solving the framing
of our theory further that the Moon is the key to the Universe. And we are going to to that now
in the next section.
of 42 71
11.0 Solving The System We now undertake since the second is Natural, to show the Earth
year is as well because the Earth goes around the Sun not just in year, but in a second in
another sense. That is we can make a variable and say:"
Equation 11.1 "
We start with"
Equation 11.2
We write
Equation 11.3
Equation 11.4
Where is the mass of the Earth, and is the mass of the Moon. We now substitute for
the mass the mass of Jupiter because it significantly carries most of the mass of the solar
system excluding the Sun. It is 1.899E27 kg. We have
Equation 11.5.
This is about 1 Earth day because 1 Earth day=86400 seconds. We want to suggest that in our
scenario that the Natural cycles in time are inversely proportional to mass:
Equation 11.6
Because if , then . If the equation is:
Equation 11.7
Where M and T are in lunar masses and Earth years.
Let . We have the mass of Jupiter in Lunar masses is
t
1
t
1
= (1secon d,1secon d /1year, . . . )
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
t
1
h
G
c
3
m
p
k
2
= 2.5721E 30kg s
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
M
e
M
m
= 2.1E 32kg s
M
e
M
m
M
e
M
j
t
1
1
M
j
h
G
c
3
m
p
k
2
M
e
M
m
=
2.1E 32kg s
1.899E 27kg
= 110584.5secon d s
1
M
T
M = M
m
1
M
m
=
1
1
= 1year = T
M
e
M
m
1
M
= T
M = M
J
M
j
=
1.899E 27kg
7.34767E 22kg
= 2.5845E4Lu n arMasses
of 43 71
This is about one day since there are 365.25 days in a year. We already said
And we have
Is about one day because there are 365.25 days in a year. Thus we equate the latter two:
But, the expression on the left gives its answer in Earth years and the expression on the right
gives the answer in seconds. Let convert seconds to Earth years. Let
We have
So since the Earth orbital period and
We have
Again we have, but we divide by to put it in lunar masses on the left and let
:
Equation 11.8.
M
e
M
m
1
M
j
=
81.27
2.5845E4
= 0.00314years
t
1
1
M
j
h
G
c
3
m
p
k
2
M
e
M
m
=
2.1E 32kg s
1.899E 27kg
= 1.2799d ays
M
e
M
m
1
M
j
=
81.27
2.5845E4
= 1.15d ays
M
e
M
m
1
M
j
= t
1
1
M
j
h
G
c
3
m
p
k
2
M
e
M
m
t
1
t
1
= (365.25)(24)(60)(60) = (1secon d )/31557600secon ds /year
t
1
1
M
j
h
G
c
3
m
p
k
2
M
e
M
m
= 110584.5secon d s
110584.5s
31557600s
= 0.00350years
31557600 = T
e
M
e
M
m
1
M
j
= 0.00314years
0.00350years 0.00314years
M
j
M
m
t
1
= 1secon d /T
e
M
e
M
m
1
M
j
/M
m
=
1secon d
T
e
1
M
j
h
G
c
3
m
p
k
2
M
e
M
m
of 44 71
This is interesting because , is not a second but the Earth orbital period in
seconds. We have
Equation 11.9
This says
Equation 11.10
Let’s verify it, we wrote
We have
The measured mass of the moon is 7.34767E22kg, We are close by
accuracy
Thus we see
Equation 11.10 is our original equation, Equation 11.2
We can write it
But now
t
1
= 1secon d /T
e
1 =
h
G
c
3
m
p
k
2
1secon d
M
m
T
e
M
m
=
h
G
c
3
m
p
k
2
1secon d
T
e
t
1
h
G
c
3
m
p
k
2
= 2.5721E 30kg s
M
m
=
2.5721E 30kg s
31557600s
= 8.15E 22kg
7.34767E 22kg
8.15E 22kg
= 90 %
t
1
= 1secon d,1secon d /year, . . .
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
h
G
c
3
m
p
M
m
(
1
k
)
2
t
1
= 1Ear thYear
t
1
= 1secon d,1secon d /1year, . . .
of 45 71
Because in lunar masses and Earth years it is to say
Since
Equation 11.11
Because
Equation 11.12
In the process we found out that
Where the mass of Jupiter gives one Earth day is one rotation of the Earth. We already had that
the mass of the Moon gives one orbit of the Earth:
We see the Solar system is like an atom, which is miraculously quantized in terms of the second,
which describes the atom and the Moon:
1
M
= T
1 =
h
G
c
3
m
p
k
2
1secon d
M
m
T
e
h
G
c
3
m
p
k
2
1second
M
m
T
e
M
= T
h
G
c
3
m
p
k
2
secon d
M
m
T
e
=
2.5721E 30kg
(7.34767E 22)(31557600s)
= 1.109 1
t
1
1
M
j
h
G
c
3
m
p
k
2
M
e
M
m
= 1Ear th Da y
h
G
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
r
p
m
p
h 4π
Gc
= 1secon d
K E
Earth
K E
moon
(1secon d ) = Ear th Da y
of 46 71
12.0 The Proton Charge
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 12.1
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
3
(x, y, z)
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
ct
2
)
i
··
x (ct
1
ct
2
)
i
x
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
ct
2
)
i
g m =
m
s
2
Fig. 2
of 47 71
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
Equation 12.2
We get
Equation 12.3 !
g =
1
s
2
=
1
t
2
1
s
2
m
s
s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
1
k
)
2
h 4πr
2
p
Gc
= q
2
h 4πr
2
p
Gc
=
(6.626E 34)(4π)(0.833E 15)
2
(6.674E 11)(299,792, 459)
= 5.37E 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E 31) = 9.1435E 19C
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71protons 6protons
of 48 71
13.0 The Solar Magnetic Field
We model the formation of the solar system from a slowly rotating gas cloud, a nebula of
gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on
and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,
like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from
which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three
factors that determine its structure, the inward gravity, the pressure gradient outward which
balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The
flattened rotating disc is broken up into rings each that has a mass spread out over it from which
the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth
masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass
spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars
masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the
Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of
about 350 because it is mostly iron condensations with incomplete silicon condensations.
Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is
-1.5 so that the density distribution of the protoplanetary disc is:
Giving a mass
With pressure gradients playing the key role in the formation of solar system, less attention is
payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of
this aspect found something very interesting concerning it. He was Alfven (1942). At the time
people were suggesting instead of the solar system forming from a rotating nebula, rather the
sun came into existence not at the same time at the center of the disc, but rather passed through
clouds and captured material after already existing. He figured for the captured material its
inward component v, and density , at a distance r from the sun, had to conserve mass, which
required:
He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions
would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the
magnetic field becomes important. He considered for simplicity the solar magnetic field was
generated by a dipole moment , a vector quantity, and that a particle moving in the plane of
that vector with mass m and charge q, would have all of both the gravitational and magnetic
forces in that plane, so the problem becomes two-dimensional and required only the and , of
polar coordinates. The differential equations of its motion would be:
σ (r) = σ
0
r
3/2
σ
0
= 3300
M =
2π
0
r
h
r
s
σ (r)r drd θ
ρ
dM
dt
= 4π r
2
ρv
μ
θ
r
of 49 71
Equation 13.1
And,
Equation 13.2
We can integrate equation 13.2 with the boundary condition that the angular momentum of the
particle is zero at large distances from the sun to get:
Equation 13.3
And substitute it into equation equation 13.1 for to get
Equation 13.4
Which we can write
Equation 13.5
We then integrate this with respect to r with the boundary condition that at large r and get
Equation 13.6
He then notices there is another value for which . It is
Equation 13.7
This is interesting because it means the particle can never approach the Sun closer than this
value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this
as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the
magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact
that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the
region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen
and helium.
Certainly today we don’t see the planets as having formed from material gathered by the Sun in
its journey, but rather think the Sun and planets formed at the same time from a cloud that
collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through
clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing
clouds from falling into their star. But we can also apply his equation to our Sun today, for which
we know a great deal about its magnetic field, which also happens to be an important thing to
study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels
m
··
r =
GM
m
r
2
+
qμ
·
θ
r
2
+ mr
·
θ
2
m
r
d(r
2
·
θ )
dt
=
qμ
·
r
r
3
mr
2
·
θ =
qμ dr
r
2
=
qμ
r
·
θ
m
··
r =
GM
r
2
+
2q
2
μ
2
m
2
r
5
·
r
d
·
r
dr
=
GM
r
2
+
2q
2
μ
2
m
2
r
5
·
r = 0
·
r
2
=
2GM
r
q
2
μ
2
m
2
r
4
·
r = 0
r
c
=
(
q
2
μ
2
2GM
m
2
)
1/3
r
c
of 50 71
with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic
field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict
solar maximums, so we have warning as to whether there will be a magnetic storm that will
knock out our electrical grid and internet, ahead of time.
During solar minimum the solar magnetic field has closed lines, that flow out one pole and into
the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so
that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The
total magnetic field of the Sun at the Earth, is all the components taken together, which are
, , and . The important component is , because it runs north-south, so it is
perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the
component that interacts with the Earth magnetosphere, and when it points southward, it will
connect with the Earth’s magnetosphere which points northward so the solar poles flow into the
Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain
down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t
always stay around the Sun itself, but the solar wind carries it through the solar system until it
collides with the interstellar medium reaching the heliopause. Thus the Sun creates the
Interplanetary Magnetic Field (IMF) which has a spiral shape because the Sun rotates once
about every 25 days. But the upshot is that at Earth we have
Moderate Magnetic Field: 10 nT
Strong Magnetic Field: 20 nT
Very Strong Magnetic Field: 30 nT
For our purposes we want to return to equation:
B
t
B
x
B
y
B
z
B
z
of 51 71
And ask just what is , because in the time that Alfven was working we worked with magnetic
fields differently, aside from his equation uses a trick, which we still use today, and that is to
consider the magnetic field a dipole. To consider it like this is to say there are two monopoles
opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,
though they are predicted by some modern theories, they have never been found. The trick is in
that by treating the North magnetic pole and South magnetic pole as separate magnetic charges
is to treat them like we do electric charges, the charge of a proton and the charge of an electron,
which can be convenient for making computations, but don’t exist that we know of. So we will
solve equation 56 for , and see what its units are so we can understand what it represents and
we will let m be the mass of a proton and q the charge of a proton. We get:
Equation 13.8
We can write these units as, by taking Coulombs (C) equal to
Equation 13.9
This is units of force per current density, which makes sense because a flowing current creates a
force. We can also write it:
Equation 13.10
Which is energy per magnetic field strength in that the SI units of magnetic field strength is
amps per meter. This tells us:
Equation 13.11
Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and
make it a proton:
We have
, , ,
r
c
=
(
q
2
μ
2
2GM
m
2
)
1/3
μ
μ
r
3
GM
m
2
p
q
2
p
= μ =
m
3
kg
C s
a m p secon ds
kg
m
s
2
m
2
a m ps
kg
m
2
s
2
m
a m ps
μ =
En erg y
Magnet icField Strength
H H
+
+ e
= 1proton = 2.18E 18J
q
p
= 1.6E 19C
m
p
= 1.67E 27kg
G = 6.67408E 11N
m
2
kg
2
M
= 1.989E 30kg
of 52 71
We want to look at the Sun as having a current flowing around its equator in a loop with its
radius
We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger
than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar
magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979E11m):
Equation 13.12
Equation 13.13
Or,…
Equation 13.14
The radius of a proton is 0.833E-15m. We have that r is:
Equation 13.15
R
= 6.957E8m
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0 a mps /m
μ =
Ioni zat ion En erg y
Magnet icField Strength
=
2.18E 18J
37.0A mps /m
= 5.892E 20
J m
A
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
r
r
p
= 5.92 6Pr oton Ra dii = carbon
of 53 71
Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it
provided for by the Sun’s magnetic field.
!
of 54 71
Let us write the computation as one equation, and verify it. We have
Where
=Ionization energy of Hydrogen
=3.47E-39 (correct)
Equation 13.16
(0.000034574)(3.47E-39)=1.19972E-43
1.19972E-43^(1/3)
=4.932E-15
So as you can see equation 7.15 is correct. It says that carbon, the basis of life is in the ratio of
the solar magnetic field and the solar gravitational field.!
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
I
=
2B
R
μ
0
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0a m ps /m
1AU = r
e
μ =
Ioni zat ion En erg y
Magnet icField Strength
=
2.18E 18J
37.0A mps /m
= 5.892E 20
J m
A
IE
H
I
1AU
=
2B
R
μ
0
1
r
e
μ
2
= (5.982E 20)
2
= 3.47E 39
μ
2
=
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
=
(2.18E 18)
2
(12.56637E 7)
2
(1.496E11)
2
4(5E 3)
2
(6.957E8)
2
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
6r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.496E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
of 55 71
Thus we have a theoretical value for the radius of the proton:
Equation 13.17
And a theoretical value for its charge
Equation 13.18
Equation 13.19
And we have the radius of a proton in terms of the solar magnetic field at Earth
Equation 13.20
All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our
constant k
Equation 13.21
And making the approximation we can with these equations eliminate in equation
5.21 using equation 5.19 in which we can eliminate k with equation 2.22.
Equation 13.22
This has an accuracy of close to 88% because in equation 7.19 the charge of six protons is
predicted by the theory to be 5.72 protons, the rest of the equations are much more accurate, but
we seek to rectify that. We write this equation so we can have an equation that defines the solar
magnetic field by solving for .
To address the accuracy of the equation for the charge of a proton, equation 7.19, we ask what is
the culprit. We suggest it is . So we solve the equation for that to see by how much it is
off. It is:
Equation 13.23
We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%
accurate, but we want to do better.
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
c
k
e
(
α
2
6
1
k
)
2
h 4πr
2
p
Gc
= q
2
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71proton s 6pr oton s
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
1 + α = 1
q
p
31850496r
2
p
=
α
4
k
e
c
4
h
π
3
G
1
(N
A
𝔼)
2
h 4π
Gc
(IE
H
)
2
μ
2
0
r
2
e
GM
m
4
p
B
2
R
2
B
(α
4
)/36
α
4
36
= q
2
k
e
c
k
2
Gc
h 4πr
2
p
of 56 71
We now eliminate on the left in equation 7.22 with equation 7.17 to find the B field of the Sun
as described by the Earth orbit as the ground state:
Equation 13.24
Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives
a magnetic field strength of 4.73E-3 Teslas. Concerning equation 7.24
We want it to be more accurate. To do that we have to substitute for , the fraction
which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy
could be in as its experimental value has the largest errors. Other possibilities are the
is right and that as well this is due to the radius of a proton having large errors, or even that it is
supposed to be the factor 8, which would be good to consider because 8-fold symmetry is very
dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear
synthesis by stars.
Because we are looking at 6 protons we had , and because of six-fold symmetry we had
and because the charge we had because of the 1/3 root on the right. So taking
and leaving the that came from substituting for k, and the as
well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation
we have:
Equation 13.25
Where
Or we can factor out all the numbers on the right the 2 and 4 in equation 7.25 and write
Equation 13.26
In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the
radius of a hydrogen atom, its ionization energy, and the solar gravitational field, with the earth
orbit as the ground state.!
r
2
p
B
2
=
2
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
GM
m
2
p
R
2
α
4
c
3
k
e
h 4π
Gc
α
4
36
= q
2
k
e
c
k
2
Gc
h 4πr
2
p
(α
4
)/36
7
4
5
r
p
(α
4
)/36
6
2
α
2
/6
q = 6q
p
6
3
N = 6
7
= 279936
3
2
/4
2
8 = 2
3
B
2
=
2
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
2GM
m
2
p
4R
2
3
2
4
2
α
4
c
3
2
3
k
e
h 4π
Gc
N = 6
7
= 279936
B
2
=
1
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
GM
m
2
p
R
2
α
4
c
3
k
e
h π
Gc
r
e
of 57 71
Protons and neutrons pack in atoms in a way that can be seen from their electron clouds. Thus
since we can say the electron cloud of helium is spherical the nucleus packing is spherical. But
since the protons and and neutrons in the nucleus are moving (But only as far as their nuclear
wall) the packing is ordered but liquid.
I believe the reason for my finds of six-fold symmetry in both the radius of a proton and it its
charge can be explained by Buckminster Fuller’s vector equilibrium. It is the most
transformable straight-line geometry if you attach sticks with flexible corners, what Fuller calls a
flex-corner. See the following illustration.
Which means it zero-frequency for omni directional closest packing of spheres is 12 spheres,
which is carbon (six protons and six neutrons) which is the basis of life as we know it. Its
diameter is then six proton radii which we found provided for by the solar magnetic field. The
frequency is the number on any symmetrically concentric shell or layer and is given by
The more protons an element has the more neutrons it needs so the strong nuclear force in the
nucleus can overcome the mutual repulsion of protons, thus holding the atoms together.
10F
2
+ 2 = 10(1)
2
+ 2 = 12 = carbon
of 58 71
Thus, to go over that of the solar magnetic field again, proposed by Hannes Alfven (1942), at the
time there was no known mechanism for it, but he suggested the relative velocity between a
neutral gas and a plasma has a critical velocity at which the gas starts to ionize and that the
atoms or molecules will not exceed this velocity until the gas becomes almost fully ionized. The
additional energy put into the system goes into ionizing the gas instead of the velocity of the
atoms, and is roughly independent of pressure and magnetic field. Critical ionization velocity
has been recognized in the laboratory for some time. It is given by equating the kinetic energy of
the atoms to the ionization potential:
Alfven found:
Gas cloud enters Solar System
A neutral atom falls into Sun due to gravity
Motion is random, collisions happen
Temperature rises
At a distance from the Sun gas will ionize
m=atomic weight
He found for a gas cloud with average voltage 12 volts. An average atomic weight of 7, the is at
Jupiter.
That is atoms fall in towards the Sun and ionize at which point the solar magnetic field pushes
them out to Jupiter orbit where a halo forms from which planets can form.
1
2
mv
2
= eV
ion
r
i
= G
Mm
eV
ion
r
i
= G
Mm
eV
ion
= 13.5E10
m
V
ion
cm
V
ion
= volts
r
i
of 59 71
Thus if we are to suggest, as we have, that solar magnetic field provides for life, whose skeletons
are the hydrocarbons, that this has something to due with the solar wind which is plasma and
we can consider it a plasma that can move in neutral gas thus being subject to Alfven’s critical
velocity. The plasma is mostly electrons, protons, alpha particles kinetic energies between 0.5
and 10 keV. There are trace amounts of heavy ions of C, N, O, Ne Mg, Is, S, and Fe. We note the
C, N, O are the most abundant elements in life chemistry. The solar wind can reach velocities of
25o000-750000 m/s. Remember the critical velocity for hydrogen is 50,900 m/s.
of 60 71
14.0 Eightfold Symmetry Since and is dimensionless, we can write in the 6 of sixfold
symmetry for it:
We see that
Equation 14.1.
Where hydrogen and carbon are the central structural elements of life composing the
hydrocarbons which are the skeletons of biological chemistry.
Thus
Is a quantity that maps things to things. We find since the orbital velocity of the moon
, we have
We have the equation
Which honed is
k v
e
= 6
1
α
2
r
p
m
p
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Da y) k v
e
1
α
2
r
p
m
p
h 4π
Gc
= 6
K E
moon
K E
earth
(Ear th Da y)
1
α
2
4πh
Gc
:
r
p
m
p
6secon d s
1
t
1
1
α
2
r
p
m
p
h 4π
Gc
= 6proton s = Carbon
1
t
6
1
α
2
r
p
m
p
h 4π
Gc
= 1proton = Hydr ogen
(
r
p
m
p
)(
1
α
2
4πh
Gc
)
=
(
4.98E11
m
kg
)(
1.21E 11kg
s
m
)
= 6secon d s
1
α
2
4πh
Gc
= 1.21E 11kg
m
s
v
m
= 1,022m /s
(1.21E 11kg
s
m
)(1022m /s) = 1.23662E 8kg
m
i
=
3
2
c
3
3
G
3
m
2
p
1/2
of 61 71
which gives gives
Equation 14.2.
Thus we have
Six-fold which is carbon (C) has map the radius to mass of a proton to the kinetic
energy of the Moon to the kinetic energy of the Earth by a factor of the Earth period of rotation,
which is 1 second and that the intermediate mass , is 8 times the factor that does that
mapping.
Equation 14.3.
Where 1 second is
Giving the radius of a proton as
From
Where is six seconds. What is the comparison of six-fold symmetry to eight-fold symmetry?
Six-fold is in the 18 groups of the periodic table that organizes elements according to their
properties. Six-fold is dynamic because 2 and 3 are the smallest prime numbers so they are the
smallest factors down to which any number can be reduced. We have
, , ,
m
i
= (67.9943)(1.01328) = 68.897kg 69kg 70kg
68.897kg
1.23662E 8kg
= 8.0865585 8
m
i
(
1
α
2
4π
Gc
v
m
)
= 8
1
6
r
p
m
p
1
α
2
h 4π
Gc
=
K E
moon
K E
earth
(Ear th Da y) = 1secon d
1
α
2
4πh
Gc
r
p
m
p
m
i
m
i
= 8
(
1
α
2
4π
Gc
v
m
)
3 2
16
h
Gπ
1
α
2
m
2
p
R
H
N
A
𝔼
= 1secon d
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
1
t
6
1
α
2
r
p
m
p
h 4π
Gc
= 1proton
t
6
2 3 = 6
3
2
= 9
2 9 = 18
3 6 = 18
of 62 71
That is dynamic. That is how you play music in 6/8 that resolves in . It is in the
Flamenco of Spain to play 3 groups of six to make 18 like in the periodic table of the elements,
then strum another group of chords six beats long usually with rhythm 2+2+2, to make 24 in
order to come around to cycles of 12. This form is called Bulerias, and is considered the most
dynamic in the genre. However, just as popular is eight, the tango flamencos, and the rumba,
which is 4+4+4+4=16. Other forms are fandangos (four groups of 3 making 12) and Sevillanas,
the same thing. Middle Eastern music has much along these lines and so does Hindu Indian
music which are North Indian Classical music (Mostly tin tal which is four groups of four to
make 16) and Ghuzals, romantic music played in 6.
12 = 2 6
of 63 71
!
of 64 71
Appendix 1 Kinetic Energies Moon and Earth
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion.
K E
moon
K E
earth
(Ear th Da y) 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E 8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 65 71
Appendix 2 Van Der Waals Radius
Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases
that behave according to kinetic-molecular theory, he described real gases which don’t. His
equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:
Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to
increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.
However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed
beyond a limit, and at close proximity to one another they interact and have a volume around
them that excludes one another, that is they have walls. He said
That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a
correction factor b. The volume of the particles is the number of particles ( ) times the volume
of one particle:
Thus there exists a sphere of radius 2r formed by two particles in contact where no other
particles can enter. It gives the correction factor
PV = n RT
V
R
= V
I
b
V
R
V
I
n
n
4
3
π r
3
of 66 71
And the volume correction for n particles is
This is the volume correction to the Ideal Gas Law. The pressure correction says real gases
exhibit less pressure because their particles interact which is a net pulling by the bulk of
particles away from the container walls.
The reduction in pressure is proportional to by a factor a. We have for reduction of pressure
that
We substitute this into the Ideal Gas Law:
This can be written as a cubic
Which allows one to compute the critical conditions of liquefaction and to derive an expression
of the principle corresponding states. In the cubic form we have as the solution three volumes
which can be used for computing the volume at and below critical temperatures.
Thus the Van Der Waals radius is estimated
For hydrogen experimentally. Therefore with
We have described the derivation of radius of a hydrogen atom from the Van Der Waals
equations that we use to get
b = (4)
4
3
π r
3
nb = 4n ×
4
3
π r
3
n
2
v
2
P
I
= P
R
+ a
n
2
V
2
(
P + a
n
2
V
2
)
(
V nb
)
= n RT
V
3
(
b +
RT
P
)
V
2
+
a
p
V
a b
P
= 0
(
π +
3
φ
2
)
(3φ 1) = 8τ : π =
P
P
c
, φ =
V
V
c
, τ =
T
T
c
4
3
π r
3
w
=
b
N
A
r
3
w
=
3
4π
b
N
A
b = 26.61
cm
3
m ol
N
A
= 6.02E 23
r
w
= 1.0967E 8cm = 1.0967E 10m
of 67 71
Where the Van Der Waals equations are
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
(
P + a
n
2
V
2
)
(
V nb
)
= n RT
V
3
(
b +
RT
P
)
V
2
+
a
p
V
a b
P
= 0
4
3
π r
3
w
=
b
N
A
of 68 71
Appendix 3 Chandrasekhar Limit
The pressure of the outer shell of star balances with the outward pressure in the core of
the star (thermal pressure). Pressure is force per unit surface area thus…
is the mass of the core pulling in the mass of the shell and is the radius of the
core. The surface area of the star is that of a sphere, . We have
The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure
times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.
The number of protons in the core is . We have
Where is the Boltzmann constant ( ). Since we must have
if the star is not to implode or explode
And we have the estimate for the temperature of the core of a star.
Fusion would not occur at the low temperature of a star like the Sun in that there would not be
enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum
mechanical tunneling. The collisions are given by the kinetic energy of the particles
. We have
The velocity v yields the minimum distance between protons as the De Broglie wavelength
P
gravit y
P =
F
A
F = m a = PA
P =
m a
A
m a = G
M
shell
M
core
r
2
core
M
core
M
shell
r
core
A = 4π r
2
core
P
gravit y
= G
M
shell
M
core
4π r
4
core
m
p
N
p
M
core
m
p
P
ther mal
=
M
core
m
p
1
4
3
π r
3
core
k
B
T
core
k
B
1.380649E 23J K
1
P
gravit y
= P
ther mal
k
B
T
core
=
1
3
GM
shell
m
p
r
core
1
2
m
2
p
v
e
2
4πϵ
0
r
min
=
1
2
m
p
v
2
of 69 71
Since the velocity is the root mean square velocity of the protons…
We have the temperature of the star is
This is another estimate. Since the mass of a star is its volume times its density
But for a star density varies with radius
If we take the derivative of both sides of the equation we have one of the equations of stellar
structure:
1.
The so-called conservation of mass equation. The force on the shell of the star is given by the
mass of the shell
Again for there to be balance gravitation pressure equals thermal pressure:
2.
Another equation of the equations of stellar structure. The so-called equation of hydrostatic
equilibrium. This can be written
If the star is an ideal gas the density of the star varies as where for a
monatomic gas and then
λ =
h
m
2
p
v
v
rms
=
3k
B
T
mp
T
min
=
(
e
2
4πϵ
0
)
2
m
p
3π
2
h
2
k
B
m =
4
3
π r
2
ρ
4π
r
0
r
2
ρ(r)dr
dm(r)
dr
= 4π r
2
ρ(r)
F
g
= G
m(r)4π r
2
ρ(r)
r
2
dr
dρ(r)
dr
= G
m(r)ρ(r )
r
2
d
dr
(
r
2
ρ(r)
dP(r)
dr
)
= 4π Gr
2
ρr
PV
γ
= consta nt
γ = 5/3
of 70 71
In stellar dynamics we write
So that
The abundance of hydrogen and helium in the universe are approximately 75% and 24%,
respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have
Ionized hydrogen and helium have and for the Sun because of high metal
content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such
pressure is given in terms of temperature (Temperature is proportional to radiation energy):
There are three kinds of pressures that can be generated by a star: gas pressure, radiation
pressure, or degeneracy pressure.
A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called
white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and
thereby collapsed under gravity but are kept from collapsing into black holes by thermal
pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a
star on the main sequence even though they are not doing fusion. It was the Indian physicist
Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity
overcome the degeneracy pressure and collapse. The non-relativistic equation is:
There are many resources available that derive this and you can find it in any textbook on
astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the
derivation to those works.
P
1
V
γ
ρ
5/3
N
V
=
ρ
μm
p
P
gas
=
ρ
μm
p
k
B
T
4 + 12
1 + 12 + 14
= 0.59
μ = 0.59
μ = 0.62
P
rad
=
4
3
σ
c
T
4
M 0.77
c
3
h
3
G
3
N
m
4
p
= 1.41
of 71 71
The Author